# 2.3 Complex numbers: roots of quadratic equations

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You probably first encountered complex numbers when you studied values of $z$ (called roots or zeros) for which the following equation is satisfied:

$a{z}^{2}+bz+c=0.$

For $a\ne 0$ (as we will assume), this equation may be written as

${z}^{2}+\frac{b}{a}z+\frac{c}{a}=0.$

Let's denote the second-degree polynomial on the left-hand side of this equation by $p\left(z\right)$ :

$p\left(z\right)={z}^{2}+\frac{b}{a}z+\frac{c}{a}.$

This is called a monic polynomial because the coefficient of the highest-power term $\left({z}^{2}\right)$ is 1. When looking for solutions to the quadratic equation ${z}^{2}+\frac{b}{a}z+$ $\frac{c}{a}=0$ , we are really looking for roots (or zeros) of the polynomial $p\left(z\right)$ . The fundamental theorem of algebra says that there are two such roots. When wehave found them, we may factor the polynomial $p\left(z\right)$ as follows:

$p\left(z\right)={z}^{2}+\frac{b}{a}z+\frac{c}{a}=\left(z-{z}_{1}\right)\left(z-{z}_{2}\right).$

In this equation, z 1 and z 2 are the roots we seek. The factored form $p\left(z\right)=$ $\left(z-{z}_{1}\right)\left(z-{z}_{2}\right)$ shows clearly that $p\left({z}_{1}\right)=p\left({z}_{2}\right)=0$ , meaning that the quadratic equation $p\left(z\right)=0$ is solved for $z={z}_{1}$ and $z={z}_{2}$ . In the process of factoring the polynomial $p\left(z\right)$ , we solve the quadratic equation and vice versa.

By equating the coefficients of ${z}^{2},{z}^{1}$ , and z 0 on the left-and right-hand sides of [link] , we find that the sum and the product of the roots ${z}_{1}$ and ${z}_{2}$ obey the equations

${z}_{1}+{z}_{2}=-\frac{b}{a}{z}_{1}{z}_{2}=\frac{c}{a}.$

You should always check your solutions with these equations.

Completing the Square. In order to solve the quadratic equation ${z}^{2}+\frac{b}{a}z+\frac{c}{a}=0$ (or, equivalently, to find the roots of the polynomial ${z}^{2}+$ $\frac{b}{a}z+\frac{c}{a}\right)$ , we “complete the square” on the left-hand side of [link] :

${\left(z+\frac{b}{2a}\right)}^{2}-{\left(\frac{b}{2a}\right)}^{2}+\frac{c}{a}=0.$

This equation may be rewritten as

${\left(z+\frac{b}{2a}\right)}^{2}={\left(\frac{1}{2a}\right)}^{2}\left({b}^{2}-4ac\right).$

We may take the square root of each side to find the solutions

${z}_{1,2}=-\frac{b}{2a}±\frac{1}{2a}\sqrt{{b}^{2}-4ac}.$

In the equation that defines the roots ${z}_{1}$ and ${z}_{2}$ , the term ${b}^{2}-4ac$ iscritical because it determines the nature of the solutions for ${z}_{1}$ and ${z}_{2}$ . In fact, we may define three classes of solutions depending on ${b}^{2}-4ac$ .

(i) Overdamped $\left({b}^{2}-4ac>0\right)$ . In this case, the roots ${z}_{1}$ and ${z}_{2}$ are

${z}_{1,2}=-\frac{b}{2a}±\frac{1}{2a}\sqrt{{b}^{2}-4ac}.$

These two roots are real, and they are located symmetrically about the point $-\frac{b}{2a}$ . When $b=0$ , they are located symmetrically about 0 at the points $±\frac{1}{2a}\sqrt{-4ac}$ . (In this case, $-4ac>0.$ ) Typical solutions are illustrated in [link] .

(ii) Critically Damped $\left({b}^{2}-4ac=0\right)$ . In this case, the roots ${z}_{1}$ and ${z}_{2}$ are equal (we say they are repeated):

${z}_{1}={z}_{2}=-\frac{b}{2a}.$

These solutions are illustrated in [link] .

(iii) Underdamped $\left({b}^{2}-4ac<0\right)$ . The underdamped case is, by far, the most fascinating case. When ${b}^{2}-4ac<0$ , then the square root in the solutions for ${z}_{1}$ and ${z}_{2}$ ( [link] ) produces an imaginary number. We may write ${b}^{2}-4ac$ as $-\left(4ac-{b}^{2}\right)$ and write ${z}_{1,2}$ as

${z}_{1,2}=-\frac{b}{2a}±\frac{1}{2a}\sqrt{-\left(4ac-{b}^{2}\right)}=-\frac{b}{2a}±j\frac{1}{2a}\sqrt{4ac-{b}^{2}}.$

These complex roots are illustrated in [link] . Note that the roots are

purely imaginary when $b=0$ , producing the result

${z}_{1,2}=±j\sqrt{\frac{c}{a}}.$

In this underdamped case, the roots ${z}_{1}$ and ${z}_{2}$ are complex conjugates:

${z}_{2}={z}_{1}^{*}.$

Thus the polynomial $p\left(z\right)={z}^{2}+\frac{b}{a}z+\frac{c}{a}=\left(z-{z}_{1}\right)\left(z-{z}_{2}\right)$ also takes the form

$p\left(z\right)=\left(z-{z}_{1}\right)\left(z-{z}_{1}^{*}\right)={z}^{2}-2Re\left[{z}_{1}\right]z+{|{z}_{1}|}^{2}.$

$Re\left[{z}_{1}\right]$ and $|{z}_{1}{|}^{2}$ are related to the original coefficients of the polynomial as follows:

$2Re\left[{z}_{1}\right]=-\frac{b}{a}$
$|{z}_{1}{|}^{2}=\frac{c}{a}$

Always check these equations.

Let's explore these connections further by using the polar representations for ${z}_{1}$ and ${z}_{2}$ :

${z}_{1,2}=r{e}^{±j\theta }.$

Then [link] for the polynomial $p\left(z\right)$ may be written in the “standard form”

$p\left(z\right)=\left(z-r{e}^{j\theta }\right)\left(z-r{e}^{-j\theta }\right)={z}^{2}-2rcos\theta z+{r}^{2}.$

$2rcos\theta =-\frac{b}{a}{r}^{2}=\frac{c}{a}$

These equations may be used to locate ${z}_{1,2}=r{e}^{±j\theta }$

$r=\sqrt{\frac{c}{a}}\theta =±{cos}^{-1}\left(\frac{-b}{\sqrt{4ac}}\right).$

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